Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int T=1048576;
int f[2222222];
long long g[1111111],s[1111111];
void C(int l,int r,int v){for(l+=T-1,r+=T+1;l^r^1;l>>=1,r>>=1)l&1?:f[l^1]=v,~r&1?:f[r^1]=v;}
int Q(int l){int s=3333333;for(l+=T;l;l>>=1)s=s<f[l]?s:f[l];return s;}
int a[1111111],b[1111111],val[1111111];
long long gmin(long long a,long long b){return a<b?a:b;}
int main()
{
	int n,k,c;
	!scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
	{
		!scanf("%d%d",&a[i],&b[i]);
		a[i]*=2;
		if(b[i]==1&&a[i]>k)return puts("-1"),0;
		s[i]=s[i-1]+a[i];
		val[i*2-1]=(k-s[i-1]%k+k-a[i]%k)%k;
		val[i*2]=(k-s[i-1]%k)%k;
	}
	val[2*n+1]=k-1;
	val[2*n+2]=0;
	sort(&val[1],&val[2*n+3]);
	c=unique(&val[1],&val[2*n+3])-val;
	for(int i=1;i<2222222;i++)f[i]=n;
	for(int i=n;i;i--)
	{
		int t=Q(lower_bound(&val[1],&val[c],(k-s[i-1]%k)%k)-val);
		if(t<n)g[i-1]=g[t]+s[t]-s[i-1]+(k-(s[t]-s[i-1])%k)%k;
		else g[i-1]=s[t]-s[i-1];
		if(b[i]==1)
		{
			int l=lower_bound(&val[1],&val[c],(k-s[i-1]%k+k-a[i]%k)%k)-val+1,r=lower_bound(&val[1],&val[c],(k-s[i-1]%k)%k)-val;
			if(l<=r)C(l,r,i-1);
			else C(l,c-1,i-1),C(1,r,i-1);
		}
	}
	long long res=233333333333333333ll;
	int lim;
	if(b[1]==1)lim=(k-a[1]%k)%k;
	else lim=k-1;
	for(int i=1;i<c&&val[i]<=lim;i++)
	{
		int t=Q(i);
		if(t<n)res=gmin(res,g[t]+s[t]+(k-(s[t]+val[i])%k)%k);
		else res=gmin(res,s[t]);
	}
	printf("%lld",res);
	return 0;
}